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ELEN [110]
3 years ago
10

One of the sides of an equilateral triangle is 32. What is the height of the triangle? Round to 2 decimal places.

Mathematics
1 answer:
Ilya [14]3 years ago
7 0
So because it's an equilateral triangle, you can cut the triangle in half and make a right triangle with a hypotenuse of 32. The bottom of the triangle would be half of 32, which is 16, because you cut the triangle in half. Now you can use the pythagorean theorem to solve for the height.

16^{2} +b^{2} =32^{2}

The answer is whatever b is:

b = \sqrt{ 32^{2} - 16^{2}

So your answer is 27.713.
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The value of b is -6.

Explanation:

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To determine the value of b, we shall solve the expression.

Applying exponent rule, \left(a^{b}\right)^{c}=a^{b c}, we get,

y^{4b}=\frac{1}{y^{24}}

Applying exponent rule, \frac{1}{a^{b}}=a^{-b}, we have,

y^{4b}=y^{-24}

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Applying this rule, we get,

4b=-24

Dividing both sides by 4, we have,

b=-6

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yarga [219]

Answer:

1.U={1,2,3,4,5}

A={2}

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2.U={1,2,3,4}

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B={2,3}

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Step-by-step explanation:

We are given that A\cap (B\cup C) and A\cup (B\cap C)

are different sets

1.We have to construct a universe set U and non empty sets A,B and C so that above set in fact the same

Suppose U={1,2,3,4,5}

A={2}

B={2,3}

C={4,5}

B\cap C=\phi

B\cup C={2,3,4,5}

A\cap (B\cup C)={2}\cap{2,3,4,5}={2}

A\cup (B\cap C)={2}\cup\phi={2}

Hence, A\cap (B\cup C)=A\cup (B\cap C)

2.We have to construct a universe set U and non empty sets A,B and C so that  above sets are in fact different

Suppose U={1,2,3,4}

A={1,2}

B={2,3}

C={4}

B\cap C=\phi

B\cup C={2,3,4}

A\cup (B\cap C)={1,2}\cup \phi={1,2}

A\cap (B\cup C)={1,2}\cap {2,3,4}={2}

Hence, A\cap (B\cup C)\neq A\cup (B\cap C)

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Answer:

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